Axial Deformations

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Jul 18, 2012 (5 years and 1 month ago)

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Strength of Materials Prof. M. S. Sivakumar












Indian Institute of Technology Madras
Axial Deformations


Introduction
Free body diagram - Revisited
Normal, shear and bearing stress
Stress on inclined planes under axial loading
Strain
Mechanical properties of materials
True stress and true strain
Poissons ratio
Elasticity and Plasticity
Creep and fatigue
Deformation in axially loaded members
Statically indeterminate problems
Thermal effect
Design considerations
Strain energy
Impact loading












Strength of Materials Prof. M. S. Sivakumar












Indian Institute of Technology Madras
1.1 Introduction
An important aspect of the analysis and design of structures relates to the deformations
caused by the loads applied to a structure. Clearly it is important to avoid deformations so
large that they may prevent the structure from fulfilling the purpose for which it is intended.
But the analysis of deformations may also help us in the determination of stresses. It is not
always possible to determine the forces in the members of a structure by applying only the
principle of statics. This is because statics is based on the assumption of undeformable,
rigid structures. By considering engineering structures as deformable and analyzing the
deformations in their various members, it will be possible to compute forces which are
statically indeterminate. Also the distribution of stresses in a given member is
indeterminate, even when the force in that member is known. To determine the actual
distribution of stresses within a member, it is necessary to analyze the deformations which
take place in that member. This chapter deals with the deformations of a structural
member such as a rod, bar or a plate under axial loading.






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Strength of Materials Prof. M. S. Sivakumar












Indian Institute of Technology Madras
1.2 Free body diagram - Revisited
The first step towards solving an engineering problem is drawing the free body diagram of
the element/structure considered.
Removing an existing force or including a wrong force on the free body will badly affect the
equilibrium conditions, and hence, the analysis.
In view of this, some important points in drawing the free body diagram are discussed
below.

Figure 1.1
At the beginning, a clear decision is to be made by the analyst on the choice of the body to
be considered for free body diagram.
Then that body is detached from all of its surrounding members including ground and only
their forces on the free body are represented.
The weight of the body and other external body forces like centrifugal, inertia, etc., should
also be included in the diagram and they are assumed to act at the centre of gravity of the
body.
When a structure involving many elements is considered for free body diagram, the forces
acting in between the elements should not be brought into the diagram.
The known forces acting on the body should be represented with proper magnitude and
direction.
If the direction of unknown forces like reactions can be decided, they should be indicated
clearly in the diagram.
Strength of Materials Prof. M. S. Sivakumar












Indian Institute of Technology Madras
After completing free body diagram, equilibrium equations from statics in terms of forces
and moments are applied and solved for the unknowns.





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Strength of Materials Prof. M. S. Sivakumar












Indian Institute of Technology Madras
1.3 Normal, shear and bearing stress
1.3.1 Normal Stress:


Figure 1.2
When a structural member is under load, predicting its ability to withstand that load is not
possible merely from the reaction force in the member.
It depends upon the internal force, cross sectional area of the element and its material
properties.
Thus, a quantity that gives the ratio of the internal force to the cross sectional area will
define the ability of the material in with standing the loads in a better way.
That quantity, i.e., the intensity of force distributed over the given area or simply the force
per unit area is called the stress.
P
A
σ
=

1.1

In SI units, force is expressed in newtons (N) and area in square meters. Consequently,
the stress has units of newtons per square meter (N/m
2
) or Pascals (Pa).
In figure 1.2, the stresses are acting normal to the section XX that is perpendicular to the
axis of the bar. These stresses are called normal stresses.
The stress defined in equation 1.1 is obtained by dividing the force by the cross sectional
area and hence it represents the average value of the stress over the entire cross section.
Strength of Materials Prof. M. S. Sivakumar












Indian Institute of Technology Madras

Figure 1.3
Consider a small area

A on the cross section with the force acting on it

F as shown in
figure 1.3. Let the area contain a point C.
Now, the stress at the point C can be defined as,
A 0
F
lim
A
∆ →

σ =


1.2

The average stress values obtained using equation 1.1 and the stress value at a point from
equation 1.2 may not be the same for all cross sections and for all loading conditions.
1.3.2 Saint - Venant's Principle:

Figure 1.4

Strength of Materials Prof. M. S. Sivakumar












Indian Institute of Technology Madras
Consider a slender bar with point loads at its ends as shown in figure 1.4.
The normal stress distribution across sections located at distances b/4 and b from one and
of the bar is represented in the figure.
It is found from figure 1.4 that the stress varies appreciably across the cross section in the
immediate vicinity of the application of loads.
The points very near the application of the loads experience a larger stress value whereas,
the points far away from it on the same section has lower stress value.
The variation of stress across the cross section is negligible when the section considered
is far away, about equal to the width of the bar, from the application of point loads.
Thus, except in the immediate vicinity of the points where the load is applied, the stress
distribution may be assumed to be uniform and is independent of the mode of application
of loads. This principle is called Saint-Venant's principle.
1.3.3 Shear Stress:

The stresses acting perpendicular to the surfaces considered are normal stresses and
were discussed in the preceding section.
Now consider a bolted connection in which two plates are connected by a bolt with cross
section A as shown in figure 1.5.

Figure 1.5
The tensile loads applied on the plates will tend to shear the bolt at the section AA.
Hence, it can be easily concluded from the free body diagram of the bolt that the internal
resistance force V must act in the plane of the section AA and it should be equal to the
external load P.
These internal forces are called shear forces and when they are divided by the
corresponding section area, we obtain the shear stress on that section.
Strength of Materials Prof. M. S. Sivakumar












Indian Institute of Technology Madras
V
A
τ
=

1.3

Equation 1.3 defines the average value of the shear stress on the cross section and the
distribution of them over the area is not uniform.
In general, the shear stress is found to be maximum at the centre and zero at certain
locations on the edge. This will be dealt in detail in shear stresses in beams (module 6).
In figure 1.5, the bolt experiences shear stresses on a single plane in its body and hence it
is said to be under single shear.
Figure 1.6
In figure 1.6, the bolt experiences shear on two sections AA and BB. Hence, the bolt is
said to be under double shear and the shear stress on each section is
V P
A 2A
τ = =

1.4

Assuming that the same bolt is used in the assembly as shown in figure 1.5 and 1.6 and
the same load P is applied on the plates, we can conclude that the shear stress is reduced
by half in double shear when compared to a single shear.
Shear stresses are generally found in bolts, pins and rivets that are used to connect
various structural members and machine components.
1.3.4 Bearing Stress:
In the bolted connection in figure 1.5, a highly irregular pressure gets developed on the
contact surface between the bolt and the plates.
The average intensity of this pressure can be found out by dividing the load P by the
projected area of the contact surface. This is referred to as the bearing stress.
Strength of Materials Prof. M. S. Sivakumar












Indian Institute of Technology Madras

Figure 1.7
The projected area of the contact surface is calculated as the product of the diameter of
the bolt and the thickness of the plate.
Bearing stress,
b
P P
A t d
σ = =
×

1.5


Example 1:

Figure 1.8
Strength of Materials Prof. M. S. Sivakumar












Indian Institute of Technology Madras
A rod R is used to hold a sign board with an axial load 50 kN as shown in figure 1.8. The
end of the rod is 50 mm wide and has a circular hole for fixing the pin which is 20 mm
diameter. The load from the rod R is transferred to the base plate C through a bracket B
that is 22mm wide and has a circular hole for the pin. The base plate is 10 mm thick and it
is connected to the bracket by welding. The base plate C is fixed on to a structure by four
bolts of each 12 mm diameter. Find the shear stress and bearing stress in the pin and in
the bolts.
Solution:
Shear stress in the pin =
(
)
( )
3
pin
2
50 10/2
V P
79.6 MPa
A 2A
0.02/4
×
τ = = = =
π

Force acting on the base plate = Pcos
θ
= 50 cos30
0
= 43.3 kN
Shear stress in the bolt,
(
)
( )
3
bolt
2
43.3 10/4
P
4A
0.012/4
×
τ = =
π

95.7 MPa=

Bearing stress between pin and rod,
(
)
( ) (
)
3
b
50 10
P
b
d 0.05 0.02
×
σ = =
× ×

50 MPa=

Bearing stress between pin and bracket =
(
)
( ) (
)
3
b
50 10/2
P/2
b
d 0.022 0.02
×
σ = =
× ×

= 56.8 MPa
Bearing stress between plate and bolts =
(
)
( ) (
)
3
b
43.3 10/4
P/4
t d 0.01 0.012
×
σ = =
× ×

90.2 MPa=


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Strength of Materials Prof. M. S. Sivakumar












Indian Institute of Technology Madras
1.4 Stress on inclined planes under axial loading:

When a body is under an axial load, the plane normal to the axis contains only the normal
stress as discussed in section 1.3.1.
However, if we consider an oblique plane that forms an angle
?
with normal plane, it
consists shear stress in addition to normal stress.
Consider such an oblique plane in a bar. The resultant force P acting on that plane will
keep the bar in equilibrium against the external load P' as shown in figure 1.9.
Figure 1.9
The resultant force P on the oblique plane can be resolved into two components F
n
and F
s

that are acting normal and tangent to that plane, respectively.
If A is the area of cross section of the bar, A/cos
?
is the area of the oblique plane. Normal
and shear stresses acting on that plane can be obtained as follows.
F
n
= Pcos
θ
F
s
= -Psin
θ
(Assuming shear causing clockwise rotation negative).
2
Pcos P
cos
A/cos A
θ
σ = = θ
θ

1.6
Psin P
sin cos
A/cos A
θ
τ = − = − θ θ
θ

1.7

Equations 1.6 and 1.7 define the normal and shear stress values on an inclined plane that
makes an angle
θ
with the vertical plane on which the axial load acts.
From above equations, it is understandable that the normal stress reaches its maximum
when
θ
= 0
o
and becomes zero when
θ
= 90
o.
Strength of Materials Prof. M. S. Sivakumar












Indian Institute of Technology Madras
But, the shear stress assumes zero value at
θ
= 0
o
and
θ
= 90
o
and reaches its maximum
when
θ
= 45
o
.
The magnitude of maximum shear stress occurring at
θ
= 45
o
plane is half of the maximum
normal stress that occurs at
θ
= 0
o
for a material under a uniaxial loading.
max
max
P
2A 2
σ
τ = =

1.8

Now consider a cubic element A in the rod which is represented in two dimension as
shown in figure 1.10 such that one of its sides makes an angle
?
with the vertical plane.

Figure 1.10
To determine the stresses acting on the plane mn, equations 1.6 and 1.7 are used as such
and to knows the stresses on plane om,
θ
is replaced by
θ
+ 90
o
.
Maximum shear stress occurs on both om and mn planes with equal magnitude and
opposite signs, when mn forms 45
o
angle with vertical plane.
Example 2:

A prismatic bar of sides 40 mm x 30 mm is axially loaded with a compressive force of 80
kN. Determine the stresses acting on an element which makes 30
0
inclination with the
vertical plane. Also find the maximum shear stress value in the bar.

Strength of Materials Prof. M. S. Sivakumar












Indian Institute of Technology Madras
Figure 1.11
Solution:

Area of the cross section, A = 40 x 30 x 10
-6
= 1.2 x 10
-3
m
2
Normal stress on 30
0
inclined plane,
2
P
cos
A
σ
= θ

3
2 o
3
80 10
cos 30 50 MPa
1.2 10

− ×
= × = −
×

Shear stress on 300 plane,
3
0 0
3
P 80 10
sin cos sin30 cos30
A
1.2 10

− ×
τ = θ θ = × ×
×

= 28.9 MPa [Counter clockwise]
Normal stress on 120
0
plane,
3
2 0
3
80 10
cos 120 16.67 MPa
1.2 10

− ×
σ = = −
×

Shear stress on 120
0
plane,
3
0 0
3
80 10
sin120 cos120 28.9 MPa
1.2 10

×
τ = × × = −
×
[Clock wise]

×
τ = =
× ×
±
3
max
3
P 80 10
Maximum shear stress in the bar,
2A
2 1.2 10
= 33.3 MPa




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Strength of Materials Prof. M. S. Sivakumar












Indian Institute of Technology Madras
1.5 Strain
The
structural member and machine components undergo deformation as they are brought
under loads.
To ensure that the deformation is within the permissible limits and do not affect the
performance of the members, a detailed study on the deformation assumes significance.
A quantity called strain defines the deformation of the members and structures in a better
way than the deformation itself and is an indication on the state of the material.

Figure 1.12
Consider a rod of uniform cross section with initial length as shown in figure 1.12.
Application of a tensile load P at one end of the rod results in elongation of the rod by
0
L
δ
.
After elongation, the length of the rod is L. As the cross section of the rod is uniform, it is
appropriate to assume that the elongation is uniform throughout the volume of the rod. If
the tensile load is replaced by a compressive load, then the deformation of the rod will be a
contraction. The deformation per unit length of the rod along its axis is defined as the
normal strain. It is denoted by
ε
0
L L

L L

δ
ε = =

1.9

Though the strain is a dimensionless quantity, units are often given in mm/mm, µm/m.


Strength of Materials Prof. M. S. Sivakumar












Indian Institute of Technology Madras
Example 3:
A circular hollow tube made of steel is used to support a compressive load of 500kN. The
inner and outer diameters of the tube are 90mm and 130mm respectively and its length is
1000mm. Due to compressive load, the contraction of the rod is 0.5mm. Determine the
compressive stress and strain in the post.
Solution
Force,
3
P 500 10 N(compressive)= − ×
Area of the tube,
( ) ( )
2 2
3 2
A 0.13 0.09 = 6.912 10 m
4

π
⎡ ⎤
= − ×
⎢ ⎥
⎣ ⎦

Stress,
3
3
P 500 10
72.3MPa (compressive)
A
6.912 10

− ×
σ = = = −
×

Strain,
4
0
0.5
5 10 (compressive)
L 1000

δ −
ε = = = − ×




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Strength of Materials Prof. M. S. Sivakumar












Indian Institute of Technology Madras
1.6 Mechanical properties of materials

A tensile test is generally conducted on a standard specimen to obtain the relationship
between the stress and the strain which is an important characteristic of the material.
In the test, the uniaxial load is applied to the specimen and increased gradually. The
corresponding deformations are recorded throughout the loading.
Stress-strain diagrams of materials vary widely depending upon whether the material is
ductile or brittle in nature.
If the material undergoes a large deformation before failure, it is referred to as ductile
material or else brittle material.









Figure 1.13
In figure 1.13, the stress-strain diagram of a structural steel, which is a ductile material, is
given.
Initial part of the loading indicates a linear relationship between stress and strain, and the
deformation is completely recoverable in this region for both ductile and brittle materials.
This linear relationship, i.e., stress is directly proportional to strain, is popularly known as
Hooke's law.
E
σ
= ε

1.10

The co-efficient E is called the modulus of elasticity or Young's modulus.
Most of the engineering structures are designed to function within their linear elastic region
only.
Strength of Materials Prof. M. S. Sivakumar












Indian Institute of Technology Madras
After the stress reaches a critical value, the deformation becomes irrecoverable. The
corresponding stress is called the yield stress or yield strength of the material beyond
which the material is said to start yielding.
In some of the ductile materials like low carbon steels, as the material reaches the yield
strength it starts yielding continuously even though there is no increment in external
load/stress. This flat curve in stress strain diagram is referred as perfectly plastic region.
The load required to yield the material beyond its yield strength increases appreciably and
this is referred to strain hardening of the material.
In other ductile materials like aluminum alloys, the strain hardening occurs immediately
after the linear elastic region without perfectly elastic region.
After the stress in the specimen reaches a maximum value, called ultimate strength, upon
further stretching, the diameter of the specimen starts decreasing fast due to local
instability and this phenomenon is called necking.
The load required for further elongation of the material in the necking region decreases
with decrease in diameter and the stress value at which the material fails is called the
breaking strength.
In case of brittle materials like cast iron and concrete, the material experiences smaller
deformation before rupture and there is no necking.

Figure 1.14

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Strength of Materials Prof. M. S. Sivakumar












Indian Institute of Technology Madras
1.7 True stress and true strain
In drawing the stress-strain diagram as shown in figure 1.13, the stress was calculated by
dividing the load P by the initial cross section of the specimen.
But it is clear that as the specimen elongates its diameter decreases and the decrease in
cross section is apparent during necking phase.
Hence, the actual stress which is obtained by dividing the load by the actual cross
sectional area in the deformed specimen is different from that of the engineering stress
that is obtained using undeformed cross sectional area as in equation 1.1
True stress or actual stress,
act
act
P
A
σ =

1.11

Though the difference between the true stress and the engineering stress is negligible for
smaller loads, the former is always higher than the latter for larger loads.
Similarly, if the initial length of the specimen is used to calculate the strain, it is called
engineering strain as obtained in equation 1.9
But some engineering applications like metal forming process involve large deformations
and they require actual or true strains that are obtained using the successive recorded
lengths to calculate the strain.
True strain
0
L
0
L
dL L
ln
L L
⎛ ⎞
= =
⎜ ⎟
⎝ ⎠


1.12

True strain is also called as actual strain or natural strain and it plays an important role in
theories of viscosity.
The difference in using engineering stress-strain and the true stress-strain is noticeable
after the proportional limit is crossed as shown in figure 1.15.
Strength of Materials Prof. M. S. Sivakumar












Indian Institute of Technology Madras

Figure 1.15



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Strength of Materials Prof. M. S. Sivakumar












Indian Institute of Technology Madras
1.8 Poissons ratio

Figure 1.16
Consider a rod under an axial tensile load P as shown in figure 1.6 such that the material
is within the elastic limit. The normal stress on x plane is
xx
P
A
σ
=
and the associated
longitudinal strain in the x direction can be found out from
xx
x
E
σ
ε =
. As the material
elongates in the x direction due to the load P, it also contracts in the other two mutually
perpendicular directions, i.e., y and z directions.
Hence, despite the absence of normal stresses in y and z directions, strains do exist in
those directions and they are called lateral strains.
The ratio between the lateral strain and the axial/longitudinal strain for a given material is
always a constant within the elastic limit and this constant is referred to as Poisson's ratio.
It is denoted by .
ν
lateral strain

axial strain
ν = −

1.13

Since the axial and lateral strains are opposite in sign, a negative sign is introduced in
equation 1.13 to make
ν
positive.
Using equation 1.13, the lateral strain in the material can be obtained by

xx
y z x
E
σ
ε = ε = −νε = −ν

1.14

Poisson's ratio can be as low as 0.1 for concrete and as high as 0.5 for rubber.
In general, it varies from 0.25 to 0.35 and for steel it is about 0.3.

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Strength of Materials Prof. M. S. Sivakumar












Indian Institute of Technology Madras
1.9 Elasticity and Plasticity
If the strain disappears completely after removal of the load, then the material is said to be
in elastic region.
The stress-strain relationship in elastic region need not be linear and can be non-linear as
in rubber like materials.

Figure 1.17
The maximum stress value below which the strain is fully recoverable is called the elastic
limit. It is represented by point A in figure 1.17.
When the stress in the material exceeds the elastic limit, the material enters into plastic
phase where the strain can no longer be completely removed.
To ascertain that the material has reached the plastic region, after each load increment, it
is unloaded and checked for residual strain.
Presence of residual strain is the indication that the material has entered into plastic
phase.
If the material has crossed elastic limit, during unloading it follows a path that is parallel to
the initial elastic loading path with the same proportionality constant E.
The strain present in the material after unloading is called the residual strain or plastic
strain and the strain disappears during unloading is termed as recoverable or elastic strain.
They are represented by OC and CD, respectively in figure.1.17.
If the material is reloaded from point C, it will follow the previous unloading path and line
CB becomes its new elastic region with elastic limit defined by point B.
Strength of Materials Prof. M. S. Sivakumar












Indian Institute of Technology Madras
Though the new elastic region CB resembles that of the initial elastic region OA, the
internal structure of the material in the new state has changed.
The change in the microstructure of the material is clear from the fact that the ductility of
the material has come down due to strain hardening.
When the material is reloaded, it follows the same path as that of a virgin material and fails
on reaching the ultimate strength which remains unaltered due to the intermediate loading
and unloading process.

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Strength of Materials Prof. M. S. Sivakumar












Indian Institute of Technology Madras
1.10 Creep and fatigue
In the preceding section, it was discussed that the plastic deformation of a material
increases with increasing load once the stress in the material exceeds the elastic limit.
However, the materials undergo additional plastic deformation with time even though the
load on the material is unaltered.
Consider a bar under a constant axial tensile load as shown in figure 1.18.

Figure 1.18
As soon as the material is loaded beyond its elastic limit, it undergoes an instant plastic
deformation at time t = 0.
0
ε
Though the material is not brought under additional loads, it experiences further plastic
deformation with time as shown in the graph in figure 1.18.
This phenomenon is called creep.
Creep at high temperature is of more concern and it plays an important role in the design
of engines, turbines, furnaces, etc.
However materials like concrete, steel and wood experience creep slightly even at normal
room temperature that is negligible.
Analogous to creep, the load required to keep the material under constant strain
decreases with time and this phenomenon is referred to as stress relaxation.
It was concluded in section 1.9 that the specimen will not fail when the stress in the
material is with in the elastic limit.
Strength of Materials Prof. M. S. Sivakumar












Indian Institute of Technology Madras
This holds true only for static loading conditions and if the applied load fluctuates or
reverses then the material will fail far below its yield strength.
This phenomenon is known as fatigue.
Designs involving fluctuating loads like traffic in bridges, and reversing loads like
automobile axles require fatigue analysis.
Fatigue failure is initiated by a minute crack that develops at a high stress point which may
be an imperfection in the material or a surface scratch.
The crack enlarges and propagates through the material due to successive loadings until
the material fails as the undamaged portion of the material is insufficient to withstand the
load.
Hence, a polished surface shaft can take more number of cycles than a shaft with rough or
corroded surface.
The number of cycles that can be taken up by a material before it fractures can be found
out by conducting experiments on material specimens.
The obtained results are plotted as
n
σ

curves as given in figure 1.19, which indicates the
number of cycles that can be safely completed by the material under a given maximum
stress.

Figure 1.19
It is learnt from the graph that the number of cycles to failure increases with decrease in
magnitude of stress.
Strength of Materials Prof. M. S. Sivakumar












Indian Institute of Technology Madras
For steels, if the magnitude of stress is reduced to a particular value, it can undergo an
infinitely large number of cycles without fatigue failure and the corresponding stress is
known as endurance limit or fatigue limit.
On the other hand, for non-ferrous metals like aluminum alloys there is no endurance limit,
and hence, the maximum stress decreases continuously with increase in number of cycles.
In such cases, the fatigue limit of the material is taken as the stress value that will allow an
arbitrarily taken number of cycles, say cycles.
8
10



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Strength of Materials Prof. M. S. Sivakumar












Indian Institute of Technology Madras
1.11 Deformation in axially loaded members
Consider the rod of uniform cross section under tensile load P along its axis as shown in
figure 1.12.
Let that the initial length of the rod be L and the deflection due to load be
δ
. Using
equations 1.9 and 1.10,
P
L E A
PL
AE
δ σ
= ε = =
δ =
E

1.15

Equation 1.15 is obtained under the assumption that the material is homogeneous and has
a uniform cross section.
Now, consider another rod of varying cross section with the same axial load P as shown in
figure 1.20.

Figure 1.20
Let us take an infinitesimal element of length dx in the rod that undergoes a deflection
d due to load P. The strain in the element is
δ
d
and d = dx
dx
δ
ε
= δ
ε

The deflection of total length of the rod can be obtained by integrating above equation,
dx
δ
= ε


L
0
Pdx
EA(x)
δ =


1.16
Strength of Materials Prof. M. S. Sivakumar












Indian Institute of Technology Madras

As the cross sectional area of the rod keeps varying, it is expressed as a function of its
length.
If the load is also varying along the length like the weight of the material, it should also be
expressed as a function of distance, i.e., P(x) in equation 1.16.
Also, if the structure consists of several components of different materials, then the
deflection of each component is determined and summed up to get the total deflection of
the structure.
When the cross section of the components and the axial loads on them are not varying
along length, the total deflection of the structure can be determined easily by,
n
i i
i i
i 1
P L
A E
=
δ =


1.17

Example 4:

Figure 1.21
Consider a rod ABC with aluminum part AB and steel part BC having diameters 25mm and
50 mm respectively as shown figure 1.21. Determine the deflections of points A and B.





Strength of Materials Prof. M. S. Sivakumar












Indian Institute of Technology Madras
Solution:


Deflection of part AB =
3
AB AB
2 9
AB AB
P L
10 10 400
A E
(0.025)/4 70 10
× ×
= −
π× × ×


0.1164 mm= −
Deflection of part BC =
3
BC BC
2 9
BC BC
P L 35 10 500
A E
(0.05)/4 200 10
× ×
= −
π× × ×


0.0446 mm= −
Deflection point of B
0.0446 mm= −
Deflection point of A
(
)
( 0.1164) 0.0446= − + −

0.161mm= −





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Strength of Materials Prof. M. S. Sivakumar












Indian Institute of Technology Madras
1.12 Statically indeterminate problems.
Members for which reaction forces and internal forces can be found out from static
equilibrium equations alone are called statically determinate members or structures.
Problems requiring deformation equations in addition to static equilibrium equations to
solve for unknown forces are called statically indeterminate problems.

Figure 1.22
The reaction force at the support for the bar ABC in figure 1.22 can be determined
considering equilibrium equation in the vertical direction.
y
F 0; R - P = 0=


Now, consider the right side bar MNO in figure 1.22 which is rigidly fixed at both the ends.
From static equilibrium, we get only one equation with two unknown reaction forces R
1
and
R
2.
1 2
- P + R + R = 0

1.18

Hence, this equilibrium equation should be supplemented with a deflection equation which
was discussed in the preceding section to solve for unknowns.
If the bar MNO is separated from its supports and applied the forces, then
these forces cause the bar to undergo a deflection
1 2
R,R and P
MO
δ
that must be equal to zero.
Strength of Materials Prof. M. S. Sivakumar












Indian Institute of Technology Madras
MO MN NO
0δ = ⇒δ + δ =
0

1.19
MN NO
and δ δ
are the deflections of parts MN and NO respectively in the bar MNO.
Individually these deflections are not zero, but their sum must make it to be zero.
Equation 1.19 is called compatibility equation, which insists that the change in length of the
bar must be compatible with the boundary conditions.
Deflection of parts MN and NO due to load P can be obtained by assuming that the
material is within the elastic limit,
1 1 2 2
MN NO
1 2
R l R l
and
A E A E
δ = δ =
.
Substituting these deflections in equation 1.19,
1 1 2 2
1 2
R l R l
- 0
A E A E
=

1.20
Combining equations 1.18 and 1.20, one can get,
1 2
1
1 2 2 1
2 1
2
1 2 2 1
PA l
R
l A l A
PA l
R
l A l A
=
+
=
+

1.21

From these reaction forces, the stresses acting on any section in the bar can be easily
determined.

Example 5:

Figure 1.23
Strength of Materials Prof. M. S. Sivakumar












Indian Institute of Technology Madras
A rectangular column of sides
0.4m 0.35m
×
, made of concrete, is used to support a
compressive load of 1.5MN. Four steel rods of each 24mm diameter are passing through
the concrete as shown in figure 1.23. If the length of the column is 3m, determine the
normal stress in the steel and the concrete. Take
steel concrete
E 200 GPa and E 29 GPa.
=
=

Solution:
P
s
= Load on each steel rod
P
c
= Load on concrete
From equilibrium equation,
c s
3
c s
P 4P P
P 4P 1.5 10..........(a)
+ =
+ = ×

Deflection in steel rod and concrete are the same.
concrete steel
δ = δ

( )
c s
9
2 9
c s
P 3 P 3
0.4 (0.35) 29 10
(0.024) 200 10
4
P 44.87P................(b)
× ×
=
π
× × ×
× × ×
=

Combining equations (a) and (b),
s
c
P 30.7kN
P 1378kN
=
=

Normal stress on concrete=
6
c
c
P
1.378 10
9.84MPa
A (0.4)(0.35)
×
= =

Normal stress on steel=
3
s
2
s
P
30.7 10
67.86MPa
A
(0.024)
4
×
= =
π



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Strength of Materials Prof. M. S. Sivakumar












Indian Institute of Technology Madras
1.13 Thermal effect
When a material undergoes a change in temperature, it either elongates or contracts
depending upon whether heat is added to or removed from the material.
If the elongation or contraction is not restricted, then the material does not experience any
stress despite the fact that it undergoes a strain.
The strain due to temperature change is called thermal strain and is expressed as
T
( T)
ε
= α ∆

1.22

where
α
is a material property known as coefficient of thermal expansion and

T indicates
the change in temperature.
Since strain is a dimensionless quantity and

T is expressed in K or
0
C,
α
has a unit that is
reciprocal of K or
0
C.
The free expansion or contraction of materials, when restrained induces stress in the
material and it is referred to as thermal stress.
Thermal stress produces the same effect in the material similar to that of mechanical
stress and it can be determined as follows.

Figure 1.24
Consider a rod AB of length L which is fixed at both ends as shown in figure 1.24.
Let the temperature of the rod be raised by

T and as the expansion is restricted, the
material develops a compressive stress.
Strength of Materials Prof. M. S. Sivakumar












Indian Institute of Technology Madras
In this problem, static equilibrium equations alone are not sufficient to solve for unknowns
and hence is called statically indeterminate problem.
To determine the stress due to

T, assume that the support at the end B is removed and
the material is allowed to expand freely.
Increase in the length of the rod
T
δ
due to free expansion can be found out using equation
1.22
T T
L ( T)δ = ε = α ∆
L

1.23

Now, apply a compressive load P at the end B to bring it back to its initial position and the
deflection due to mechanical load from equation 1.15,
T
PL
AE
δ =

1.24

As the magnitude of and
δ
are equal and their signs differ,
T
δ
T
PL
( T)L
AE
δ = −δ
α ∆ = −

T
P
Thermal stress, ( T)E
A
σ = = −α ∆

1.25

Minus sign in the equation indicates a compressive stress in the material and with
decrease in temperature, the stress developed is tensile stress as

T becomes negative.
It is to be noted that the equation 1.25 was obtained on the assumption that the material is
homogeneous and the area of the cross section is uniform.
Thermoplastic analysis assumes significance for structures and components that are
experiencing high temperature variations.






Strength of Materials Prof. M. S. Sivakumar












Indian Institute of Technology Madras
Example 6:

Figure 1.25
A rod consists of two parts that are made of steel and aluminum as shown in figure 1.25.
The elastic modulus and coefficient of thermal expansion for steel are 200GPa and 11.7 x
10
-6
per
0
C respectively and for aluminum 70GPa and 21.6 x 10
-6
per
0
C respectively. If the
temperature of the rod is raised by 50
0
C, determine the forces and stresses acting on the
rod.
Solution:
Deflection of the rod under free expansion,
( ) (
T
6 6
( T)L
11.7 10 50 500 21.6 10 50 750
1.1025 mm
− −
δ = α ∆
= × × × + × × ×
=
)

Restrained deflection of rod = 1.1025 - 0.4 = 0.7025 mm
Let the force required to make their elongation vanish be R which is the reaction force at
the ends.
Strength of Materials Prof. M. S. Sivakumar












Indian Institute of Technology Madras
[ ]
[ ]
steel Al
2
3 2
2
3 2
3 9 3
RL RL
AE AE
Area of steel rod = 0.05 1.9635 10 m
4
Area of aluminium rod = 0.03 0.7069 10 m
4
500 750
0.7025 R
1.9635 10 200 10 0.7069 10 70 10
Compressive force on the r


− −
⎛ ⎞ ⎛ ⎞
−δ = +
⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
π
= ×
π
= ×
⎡ ⎤
− = +
⎢ ⎥
× × × × × ×⎣ ⎦
9
3
3
3
3
od, R = 42.076 kN
P 42.76 10
Compressive stress on steel, = 21.8MPa
A
1.9635 10
P 42.76 10
Compressive stress on steel, = 60.5MPa
A
0.7069 10



− ×
σ = = −
×
− ×
σ = = −
×






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Strength of Materials Prof. M. S. Sivakumar












Indian Institute of Technology Madras
1.14. Design considerations
:
A good design of a structural element or machine component should ensure that the
developed product will function safely and economically during its estimated life time.
The stress developed in the material should always be less than the maximum stress it
can withstand which is known as ultimate strength as discussed in section 1.6.
During normal operating conditions, the stress experienced by the material is referred to as
working stress or allowable stress or design stress.
The ratio of ultimate strength to allowable stress is defined as factor of safety.

Ultimatestress
Factor of safety
Allowablestress
=

1.26

Factor of safety can also be expressed in terms of load as,


Ultimateload
Factor of safety
Allowableload
=

1.27

Equations 1.26 and 1.27 are identical when a linear relationship exists between the load
and the stress.
This is not true for many materials and equation 1.26 is widely used in design analysis.
Factor of safety take care of the uncertainties in predicting the exact loadings, variation in
material properties, environmental effects and the accuracy of methods of analysis.
If the factor of safety is less, then the risk of failure is more and on the other hand, when
the factor of safety is very high the structure becomes unacceptable or uncompetitive.

Hence, depending upon the applications the factor of safety varies. It is common to see
that the factor of safety is taken between 2 and 3.
Stresses developed in the material when subjected to loads can be considered to be
uniform at sections located far away from the point of application of loads.
This observation is called Saint Venant’s principle and was discussed in section 1.3.
But, when the element has holes, grooves, notches, key ways, threads and other
abrupt changes in geometry, the stress on those cross-sections will not be uniform.
Strength of Materials Prof. M. S. Sivakumar












Indian Institute of Technology Madras
These discontinuities in geometry cause high stresses concentrations in small regions of
the material and are called stress raisers.
Experimentally it was found that the stress concentrations are independent of the material
size and its properties, and they depend only on the geometric parameters.


Figure 1.26

Consider a rectangular flat plate with a circular hole as shown in figure 1.26.
The stress distribution on the section passing through the centre of the hole indicates that
the maximum stress occurs at the ends of the holes and it is much higher than the average
stress.
Since the designer, in general, is more interested in knowing the maximum stress rather
than the actual stress distribution, a simple relationship between the in
terms of geometric parameters will be of practical importance.
max ave
andσ σ
Many experiments were conducted on samples with various discontinuities and the
relationship between the stress concentration factor and the geometrical parameters are
established, where

Stress concentration factor,
max
ave
K
σ
=
σ

1.28

Strength of Materials Prof. M. S. Sivakumar












Indian Institute of Technology Madras
Hence, simply by calculating the average stress,
ave
P
A
σ
=
, in the critical section of a
discontinuity, can be easily found and by multiplying
max
σ
ave
σ
with K.
The variation of K in terms of r/d for the rectangular plate with a circular hole is given in
figure 1.26.
It is to be noted that the expression in equation 1.28 can be used as long as is within
the proportional limit of the material.
max
σ

Example 7:



Figure 1.27
A rectangular link AB made of steel is used to support a load W through a rod CD as
shown in figure 1.27. If the link AB is 30mm wide, determine its thickness for a factor of
safety 2.5. The ultimate strength of steel may be assumed to be 450 MPa.
Solution:

Drawing free body diagram of the link and the rod,




Taking moment about C,

Strength of Materials Prof. M. S. Sivakumar












Indian Institute of Technology Madras
0
BA
BA
BA
6
a
3
6
25 550 F sin60 500 0
F 31.75 kN
Tensionalonglink AB, F 31.75 kN.
Ultimatestress
F.O.S
allowablestress
450 10
Allowablestressinlink AB,180MPa
2.5
Tensileforce
Stressinlink AB,
Area
31.75 10
180 10
0.03
× − × =
=
=
=
×
σ = =
σ =
×
× =
t
Thicknessof link AB,t 5.88
t 6mm
×
=





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Strength of Materials Prof. M. S. Sivakumar












Indian Institute of Technology Madras
1.15. Strain energy:

Strain energy is an important concept in mechanics and is used to study the response of
materials and structures under static and dynamic loads.
Within the elastic limit, the work done by the external forces on a material is stored as
deformation or strain that is recoverable.
On removal of load, the deformation or strain disappears and the stored energy is
released. This recoverable energy stored in the material in the form of strain is called
elastic strain energy.



Figure 1.28
Consider a rod of uniform cross section with length L as shown in figure 1.28.
An axial tensile load P is applied on the material gradually from zero to maximum
magnitude and the corresponding maximum deformation is δ.
Area under the load-displacement curve shown in figure 1.28 indicates the work done on
the material by the external load that is stored as strain energy in the material.
Let dW be the work done by the load P due to increment in deflection dδ. The
corresponding increase in strain energy is dU.
When the material is within the elastic limit, the work done due to dδ,

dW dU Pd= =
δ

The total work done or total elastic strain energy of the material,


0
W U Pd
δ
=
=
δ


1.29

Strength of Materials Prof. M. S. Sivakumar












Indian Institute of Technology Madras
Equation 1.29 holds for both linear elastic and non-linear elastic materials.
If the material is linear elastic, then the load-displacement diagram will become as shown
in figure 1.29.



Figure 1.29

The elastic strain energy stored in the material is determined from the area of triangle
OAB.

1 1
1
U P
2
=
δ

1.30

1
1
P L
where
AE
δ =
.

Since the load-displacement curve is a straight line here, the load can be expressed in
terms of stiffness and deflection as
1
P
1
P k
1
=
δ
. Then equation 1.30 turns out to be,

2
1
1
U k
2
=
δ

1.31

Work done and strain energy are expressed in N-m or joules (J).
Strain energy defined in equation 1.29 depends on the material dimensions.
In order to eliminate the material dimensions from the strain energy equation, strain energy
density is often used.
Strain energy stored per unit volume of the material is referred to as strain energy density.
Dividing equation 1.29 by volume,

Strength of Materials Prof. M. S. Sivakumar












Indian Institute of Technology Madras
Strain energy density,

0
u
d
ε
=
σ ε


1.32

Equation 1.32 indicates the expression of strain energy in terms of stress and strain, which
are more convenient quantities to use rather than load and displacement.



Figure 1.30
Area under the stress strain curve indicates the strain energy density of the material.
For linear elastic materials within proportional limit, equation 1.32 gets simplified as,


Strain energy density,

1 1
1
u
2
=
σ ε

1.33

Using Hook’s law,
1
1
E
σ
ε =
, strain energy density is expressed in terms of stress,

2
1
u
2E
σ
=

1.34

When the stress in the material reaches the yield stress
y
σ
, the strain energy density
attains its maximum value and is called the modulus of resilience.


Modulus of resilience,
2
Y
R
u
2E
σ
=

1.35

Strength of Materials Prof. M. S. Sivakumar












Indian Institute of Technology Madras
Modulus of resilience is a measure of energy that can be absorbed by the material due to
impact loading without undergoing any plastic deformation.


Figure 1.31

If the material exceeds the elastic limit during loading, all the work done is not stored in the
material as strain energy.
This is due to the fact that part of the energy is spent on deforming the material
permanently and that energy is dissipated out as heat.
The area under the entire stress strain diagram is called modulus of toughness, which is a
measure of energy that can be absorbed by the material due to impact loading before it
fractures.
Hence, materials with higher modulus of toughness are used to make components and
structures that will be exposed to sudden and impact loads.

Example 8:


Figure 1.32
Strength of Materials Prof. M. S. Sivakumar












Indian Institute of Technology Madras
A 25 kN load is applied gradually on a steel rod ABC as shown in figure 1.32. Taking
E=200 GPa, determine the strain energy stored in the entire rod and the strain energy
density in parts AB and BC. If the yield strength of the material is σ
y
=320MPa, find the
maximum energy that can be absorbed by the rod without undergoing permanent
deformation.
Solution:



( )
2
AB
AB
2
3
AB
9
2
3
Strain energy density in part AB, u
2E
1 25 10
u
2 200 10
0.024
4
= 7.63 kJ/m
σ
=




×
=


π
× ×





( )
2
BC
BC
2
3
BC
9
2
3
Strain energy density in part BC, u
2E
1 25 10
u
2 200 10
0.016
4
= 38.65 kJ/m
σ
=




×
=


π
× ×






Strain energy in the entire rod,

( )
( )
AB AB BC BC
2 2
3 3
U u V u V
7.63 10 0.024 1 38.65 10 0.016 0.8
4 4
U 9.67J
= +
Π Π
⎡ ⎤ ⎡
= × × × + × × ×
⎢ ⎥ ⎢
⎣ ⎦ ⎣
=





The load that will produce yield stress in the material,
( )
2
6
y BC
P A 320 10 0.016
4
P 64.3kN
Π
= σ = × ×
=


Maximum energy that can be stored in the rod
,

Strength of Materials Prof. M. S. Sivakumar












Indian Institute of Technology Madras
( )
( )
( )
22
AB BC
AB BC
2
BCAB
AB BC
2
3
9
2 2
1 P P
U V V
2E A A
LLP
2E A A
64.3 10
1 0.8
2 200 10
0.024 0.016
4 4
63.97J
⎡ ⎤
⎛ ⎞
⎛ ⎞
⎢ ⎥
= × + ×
⎜ ⎟⎜ ⎟
⎢ ⎥
⎝ ⎠
⎝ ⎠
⎣ ⎦
⎡ ⎤
= +
⎢ ⎥
⎣ ⎦
⎡ ⎤
×
⎢ ⎥
= +
⎢ ⎥
Π Π
× ×
⎢ ⎥
⎣ ⎦
=



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Strength of Materials Prof. M. S. Sivakumar












Indian Institute of Technology Madras
1.16 Impact loading
A static loading is applied very slowly so that the external load and the internal force are
always in equilibrium. Hence, the vibrational and dynamic effects are negligible in static
loading.
Dynamic loading may take many forms like fluctuating loads where the loads are varying
with time and impact loads where the loads are applied suddenly and may be removed
immediately or later.
Collision of two bodies and objects freely falling onto a structure are some of the examples
of impact loading.
Consider a collar of mass M at a height h from the flange that is rigidly fixed at the end of a
bar as shown in figure 1.33.
As the collar freely falls onto the flange, the bar begins to elongate causing axial stresses
and strain within the bar
.


Figure 1.33
After the flange reaching its maximum position during downward motion, it moves up due
to shortening of the bar.
The bar vibrates in the axial direction with the collar and the flange till the vibration dies out
completely due to damping effects.
To simplify the complex impact loading analysis, the following assumptions are made.
Strength of Materials Prof. M. S. Sivakumar












Indian Institute of Technology Madras
The kinetic energy of the collar at the time of striking is completely transformed into strain
energy and stored in the bar.
But in practice, not all the kinetic energy is stored in the material as some of the energy is
dissipated out as heat and waves.
Hence, this assumption is conservative in the sense that the stress and deflection
predicted by this way are higher than the actual values.
The second assumption is that after striking the flange, the collar and the flange move
downward together without any bouncing.
This assumption is reasonable provided the weight of the collar is much larger than that of
the bar.
The third assumption is that the stresses in the bar remain within linear elastic range and
the stress distribution is uniform within the bar.
But, in reality, the stress distribution is not uniform since the stress waves generated due
to impact loading travel through the bar.
Using the principle of conservation of energy, the kinetic energy of the collar is equated to
the strain energy of the bar.
Assuming the height of fall h is much larger than the deformation of rod, and using
equation 1.34,

2
2
max
1

2 2
V
Mv
E
σ
=

1.36

where v is the velocity of the collar at strike (
2ghv =
) and V is the volume of the material.
The maximum stress in the bar due to the impact load of mass M,

2
max
M
v E
V
σ =

1.37

From above equation, it becomes clear that by increasing the volume of material, the effect
of impact loading can be minimized.
Expressing strain energy in terms of deflection in equation 1.36,
2
2
max
1

2 2
EA
Mv
L
δ
=

Strength of Materials Prof. M. S. Sivakumar












Indian Institute of Technology Madras

2
max
M
v L
E
A
δ =

1.38

If the load of the collar is applied gradually on the bar i.e., under static loading, the static
deflection
δ
st
will be,
st
MgL
EA
δ =


Substituting this in equation 1.38, relationship between the static deflection
δ
st
and the
impact deflection
δ
max
is obtained.

max st
2h
δ
= δ

1.39

To represent the magnification of deflection due to impact load compared to that of static
deflection for the same load, impact factor is used.

max
st
Impact factor
δ
=
δ

1.40

Alternately, the impact factor can be obtained from the ratio
max
st
σ
σ
.
The relationship between the stress
st
σ
developed in the bar due to static loading and the
impact loading stress is determined as follows.
max
σ

st
st st
max
max max
max st
E
= E =
L
E
= E =
L
E
= 2h
L
δ
σ ε
δ
σ ε
σ δ


st
max
2hE
=
L
σ
σ

1.41

Now, the effect of suddenly applied loads on materials or structures that forms a special
case of impact loading is discussed.
Strength of Materials Prof. M. S. Sivakumar












Indian Institute of Technology Madras
In figure 1.33, if the collar is brought into contact on the top of the flange and released
immediately, it is referred to as suddenly applied load.
The maximum stress produced in the bar due to suddenly applied load can be determined
by replacing h by
δ
max
in equation 1.41
max st
= 2
σ
σ

1.42

Hence, the stress developed in a material due to suddenly applied load is twice as large as
that of gradually applied load.

Example 9:



Figure 1.34
A 50 kg collar is sliding on a cable as shown in figure 1.34 from a height h = 1m. Its free
fall is restrained by a stopper at the end of the cable. The effective cross-sectional area
and the elastic modulus of the cable are taken to be 60 mm
2
and 150GPa respectively. If
the maximum allowable stress in the cable due to impact load is 450MPa, calculate the
minimum permissible length for the cable and the corresponding maximum deflection. Also
find the impact factor.
Solution:
Maximum stress due to impact load,

2
max
M
=
v E
V
σ


Strength of Materials Prof. M. S. Sivakumar












Indian Institute of Technology Madras

Velocity, v
=
2gh = 2 9.81 1 = 4.43 m/s× ×



( )
( )
2
9
6
6
50 4.43 150 10
450 10 =
60 10 L

× × ×
×
× ×


Minimum permissible length for the cable, L = 12.1 m


Static deflection,

st
MgL
EA
δ =

9 6
50 9.81 12.1
150 10 60 10

×
×
=
×
× ×


= 0.656 mm

Maximum deflection,
max st
2h
δ
= δ


2 1000 0.656
36mm
= × ×
=

Impact factor,

max
st
36
0.656
δ
= =
δ


= 55






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